Let $A$ be an $n \times n$ matrix of rank r. Then by reordering the indices if necessary we can bring the matrix in the form $(\frac{A_1}{A_2})$ where $A_1$ is an $r \times n$ matrix, $A_2$ is an $n-r \times n$ matrix and $A_1$ has rank r.

similarly by reordering the indices we can write the matrix $A$ as a direct sum of in decomposable matrices.

I am just thinking how to prove these two facts.

{{{{{ Note: The term irreducible is usually used instead of indecomposable.

Wikipedia: "...a matrix is irreducible if it is not similar via a permutation to a block upper triangular matrix (that has more than one block of positive size)." (Replacing non-zero entries in the matrix by one, and viewing the matrix as the adjacency matrix of a directed graph, the matrix is irreducible if and only if the digraph is irreducible.)

PlanetMath: reducible matrix "An n×n matrix A is said to be a reducible matrix if and only if for some permutation matrix P, the matrix PTAP is block upper triangular matrix." If a square matrix is not reducible, it is said to be an irreducible matrix }}}}}

When we reordering the indices actually what we are doing geometrically or algebraically about the matrix?

Thanks for your valuable time and interest.


You just need to choose $r$ linearly independent rows and move them to the top. That would be left-multiplying by an $n \times n$ permutation matrix.

  • $\begingroup$ nice. similarly how to bring $A$ into block diagonal form by reordering the indices? $\endgroup$ – GA316 May 27 '15 at 3:00
  • $\begingroup$ @GA316 What do you mean by block diagonal form? Do you mean you want an invertible $r \times r$ submatrix in the top left? If so, choose $r$ columns such that those columns of $A_1$ are linearly independent, and move them to the left. Do so by multiplying on the right by a permutation matrix. $\endgroup$ – Eric Auld May 27 '15 at 3:05
  • $\begingroup$ I have made some edit to the question. please see it. $\endgroup$ – GA316 May 27 '15 at 3:55
  • $\begingroup$ by reordering the indices I mean that we have to apply a permutation to rows as well as to columns also. but in your answer we are applying the permutation only to the rows. if we apply the same permutation to the columns also, still we get the matrix in required form? $\endgroup$ – GA316 Sep 1 '15 at 10:35

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